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Am I correct to assume that this system of differential equations only has one equilibrium $(I_{1}, I_{2}, Q_{1}, Q_{2})$, namely: $(0,0,0,0)$ ? Would we also then say this is a stable solution?

$$ \frac{d Q_{1}}{d t} = I_{1} \\ \frac{d Q_{2}}{d t} = I_{2} \\ \frac{d I_{1}}{d t} = \frac{\frac{0.2 \, * \, 10^{-6}}{10^{-7}} \, * Q_{2} - \frac{L}{10^{-7}} \, * \, Q_{1}}{{L}^2 \, - \, 0.04 \, * \, 10^{-12}} \\ \frac{d I_{1}}{d t} = \frac{\frac{0.2 \, * \, 10^{-6}}{10^{-7}} \, * Q_{1} - \frac{L}{10^{-7}} \, * \, Q_{2}}{{L}^2 \, - \, 0.04 \, * \, 10^{-12}} \\ Q_{1}(0) = 0.2 \\ Q_{2}(0) = 0 \\ I_{1}(0) = 0 \\ I_{2}(0) = 0 $$

Nick Shark
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1 Answers1

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Define $$x =\begin{bmatrix} I_1\\ I_2\\ Q_1\\ Q_2\end{bmatrix}$$

and write your system of equations as $$ \frac{dx}{dt} = A x, $$ where $A$ is a coefficient matrix. If $A$ is invertible, then the zero solution is the only possible solution of $Ax=0$. Otherwise, $\frac{dx}{dt} =0$ may have other solutions, all in the null-space of $A$. These solutions are strictly stable iff $A$ is a Hurwitz matrix, i.e., all its eigenvalues have negative real parts. So for a given value of $L$ you may check the eigenvalues of $A$ and see whether the zero solution is the only stable solution. It may be quite tricky to prove that the system is stable for all values of $L$, though Gershgoring’s circles might help.

Amit Hochman
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